Question

Consider the titration of \(50.0 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HA}\left(K_{\mathrm{a}}=\right.\) \(1.0 \times 10^{-4}\) ) with 0.010 M NaOH. (a) Sketch the pH titration curve, and label the equivalence point. (b) How many milliliters of \(0.010 \mathrm{M} \mathrm{NaOH}\) are required to reach the equivalence point? (c) Is the \(\mathrm{pH}\) at the equivalence point greater than, equal to, or less than \(7 ?\) (d) What is the pH exactly halfway to the equivalence point?

Short answer

(b) 50.0 mL of NaOH is required for equivalence. (c) pH > 7 at the equivalence point. (d) pH is 4.00 halfway to equivalence.

Step-by-step solution

Analyze the Chemical Reaction

The titration involves a weak acid (HA) being titrated by a strong base (NaOH). The reaction is:\[\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\]The equivalence point occurs when the moles of HA equal the moles of OH-, resulting in conversion to the conjugate base, A-.

Calculate Moles of HA

We start by calculating the moles of HA present initially:\[\text{moles of HA} = \text{volume (L)} \times \text{concentration (M)} = 0.050 \times 0.010 = 5.0 \times 10^{-4} \text{ moles}\]

Determine Volume of NaOH at Equivalence Point

At the equivalence point, the moles of NaOH added equal the initial moles of HA. Since the concentration of NaOH is the same as HA (0.010 M), the volume of NaOH required is:\[\text{Volume of NaOH (mL)} = \frac{5.0 \times 10^{-4} \text{ moles}}{0.010 \text{ M}} = 50.0 \text{ mL}\]

Assess pH at Equivalence Point

The equivalence point for the titration of a weak acid with a strong base will have a pH greater than 7 because the solution will consist of the conjugate base A- in water. A- will hydrolyze:\[\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-\]This process increases the pH.

Calculate pH Halfway to Equivalence Point

Halfway to the equivalence point, the amount of HA equals the amount of A-. This is the point where:\[\text{pH} = \text{pK}_a\]Given \( K_a = 1.0 \times 10^{-4} \), we calculate \( pK_a \):\[pK_a = -\log(1.0 \times 10^{-4}) = 4.00\]Thus, the pH halfway is 4.00.

Key concepts

Titration Curve

A titration curve is a graphical representation of the change in pH as a titrant is added to a solution. During a titration, the curve typically starts at the initial pH of the solution being titrated. The curve will gradually rise as the titrant is added and can show different shapes based on whether the substance is a strong or weak acid or base.

For the exercise provided, since we are titrating a weak acid (HA) with a strong base (NaOH), the initial pH is below 7. As NaOH is added, the pH increases, reflecting the neutralization of the acid. This results in a titration curve that increases smoothly until it nears the equivalence point, where the slope becomes steeper.

• Before the equivalence point: The solution predominantly contains the weak acid along with its partially neutralized form.
• At the equivalence point: The weak acid is fully neutralized, forming its conjugate base.
• Beyond the equivalence point: The solution is basic as excess hydroxide ions (OH-) are present.

Sketching this curve helps identify the equivalence point where the moles of titrant equal the moles of the analyte.

Equivalence Point

The equivalence point in titration is where the amount of titrant added exactly neutralizes the analyte in the solution. For a titration involving a weak acid and a strong base, such as our case where HA is titrated with NaOH, the equivalence point is crucial.

At this stage, all the weak acid (HA) is converted into its conjugate base (A-), and the solution consists primarily of this base. This results in a basic pH, typically greater than 7, due to the hydrolysis of the conjugate base. The equivalence point is not necessarily at pH 7, which is reserved for strong acid-strong base reactions. Instead, here, it reveals the strength and behavior of the conjugate base in water.

In the case of the exercise, the equivalence point is reached after adding 50.0 mL of NaOH, where the moles of NaOH equal the moles of HA initially present.

Weak Acid

A weak acid, like HA in our exercise, partially ionizes in solution. This is in contrast to strong acids which completely dissociate in water. The degree of ionization of a weak acid is represented by its acid dissociation constant, \(K_a\).

With a given \(K_a = 1.0 \times 10^{-4}\), the weak acid HA only partially dissociates to donate protons (H+) in water, and this affects both the initial pH of the solution and the pH changes during titration.

• The lower the \(K_a\), the weaker the acid.
• A weak acid results in a more gradual increase in a titration curve prior to reaching the equivalence point.
• Its conjugate base, formed at the equivalence point, can cause the pH to be greater than 7.

Understanding the properties of weak acids is crucial in predicting the behavior of a titration curve and calculating the pH at various stages of the process.

pH Calculation

Calculating the pH during a titration involves understanding the chemical reactions taking place, the concentrations of all substances involved, and sometimes, the use of the Henderson-Hasselbalch equation.

In the context of the exercise, the pH calculation differs depending on whether you're before, at, or after the equivalence point. Halfway to the equivalence point is a special scenario for weak acid-strong base titrations, where the pH is equal to the \(pK_a\) of the weak acid.

Since \(K_a = 1.0 \times 10^{-4}\), the \(pK_a\) can be calculated as follows:\[pK_a = -\log(1.0 \times 10^{-4}) = 4.00\]Hence, at the halfway point of the titration, the pH is 4.00. This is because the concentrations of the acid (HA) and its conjugate base (A-) are equal, simplifying the process of pH calculation.

Accurate pH calculations help in predicting how much titrant is needed to achieve desired pH changes in any titration process.